Question

If ${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=\pi \left({\sec }^2\right(u)+{\sec }^4(v)+{\sec }^6(w\left)\right)$, where $u,v,w$ are least non-negative angles such that $u<v<w$, then the value of $x^{2000}+y^{2002}+z^{2004}+\frac{36\pi }{u+v+w}$ is

• A. $3\pi$
• B. $3$
• C. $9\pi$
• D. $9$

Answer 1

The correct option is D $9$

${\sec }^2\left(u\right),{\sec }^4\left(v\right),{\sec }^6\left(w\right)\in (1,\infty )$
$\Rightarrow {\sec }^2\left(u\right)+{\sec }^4\left(v\right)+{\sec }^6\left(w\right)\in (3,\infty )$
$\Rightarrow \pi \left({\sec }^2\right(u)+{\sec }^4(v)+{\sec }^6(w\left)\right)\in (3\pi ,\infty )$
Similarly
${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z\in (0,3\pi )$
$\therefore$ The equation is true only if it is equal to $3\pi$
$\Rightarrow {\cos }^{-1}x={\cos }^{-1}y={\cos }^{-1}z=\pi \Rightarrow x=y=z=-1$ and ${\sec }^{2}u={\sec }^{4}v={\sec }^{6}w=1$
$\Rightarrow u=\pi ,v=2\pi ,w=3\pi \Rightarrow x^{2000}+y^{2002}+z^{2004}+\frac{26\pi }{u+v+w}=1+1+1+\frac{36\pi }{6\pi }=9$